Differentiating Logarithmic Functions with Base e

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Differential Calculus #8: Logarithmic Differentiation

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Darshan Senthil

Key Questions

  • Answer:

    #1/x#

    Explanation:

    #log_e(x)# is commonly denoted as #ln(x)#, the natural log.

    #=>d/(dx) ln(x) = 1/x#

    If you would like a proof, we can derive it from the limit definition:

    #lim_(delta x->0)(f(x+delta x)-f(x))/(delta x)#

    #= lim_(delta x->0)(ln(x+delta x)-ln(x))/(delta x)#

    #= lim_(delta x->0)(ln((x+delta x)/(x)))/(delta x)#

    #= lim_(delta x->0)1/(delta x)ln(1+(delta x)/x)#

    #= lim_(delta x->0)ln((1+(delta x)/x)^(1/(delta x)))#

    #= lim_(delta x->0)ln((1+(delta x)/x)^(1/(delta x)))#

    #"Let " tau equiv (delta x)/x#:

    #= lim_(delta tau->0)ln((1+tau)^(1/(xtau)))#

    #= lim_(delta tau->0)ln[((1+tau)^(1/(tau)))^(1/x)]#

    #= ln[(e)^(1/x)]#

    #= 1/x ln(e)#

    #= 1/x (1)#

    #= 1/x#

  • By Quotient Rule,

    #y'={1/x cdot x-lnx cdot 1}/{x^2}={1-lnx}/{x^2}#

    This problem can also be solved by the Product Rule

    #y'=f'(x)g(x)+f(x)g(x)#

    The original function can also be rewritten using negative exponents.

    #f(x)=ln(x)/x=ln(x)*x^-1#

    #f'(x)=1/x*x^-1+ln(x)*-1x^-2#

    #f'(x)=1/x*1/x+ln(x)*-1/x^2#

    #f'(x)=1/x^2-ln(x)/x^2#

    #f'(x)=(1-ln(x))/x^2#

  • The answer would be #f'(x) = 1/g(x)*g'(x)# or it can be written as #f'(x)=(g'(x))/g(x)#.

    To solve this derivative you will need to follow the chain rule which states:

    #F(x)=f(g(x))# then #F'(x)=f'(g(x))*g'(x)#

    Or without the equation, it the derivative of the outside(without changing the inside), times the derivative of the outside.

    The derivative of #h(x) = ln(x)# is #h'(x) = 1/x#.

Questions

  • Double-check the answer
    Henry W. answered · 5 months ago

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Differentiating Logarithmic Functions